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Can anyone explain why a semi-simple finite dimensional Lie algebra $\mathfrak{g}$ has to be perfect ?

The natural way to prove something like that would be to look to the algebra generated by the Lie brackets, which when $\mathfrak{g}$ is not perfect would be expected to be solvable. But it doesn't seem to work.

Selim Ghazouani
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4 Answers4

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Let $\mathfrak{g}$ be a semi simple Lie algebra. Then $\mathfrak{g}$ can be written as direct sum of simple ideals.

For simplicity, we can write $\mathfrak{g}=\mathfrak{g}_1\oplus\mathfrak{g}_2$ where $\mathfrak{g}_1,\mathfrak{g}_2$ are simple ideals.

We then have $$[\mathfrak{g},\mathfrak{g}]=[\mathfrak{g}_1\oplus \mathfrak{g}_2, \mathfrak{g}_1\oplus \mathfrak{g}_2]= [\mathfrak{g}_1,\mathfrak{g}_1]\oplus [\mathfrak{g}_2,\mathfrak{g}_2]=\mathfrak{g}_1\oplus \mathfrak{g}_2=\mathfrak{g}$$

See that $\mathfrak{g}_1=[\mathfrak{g}_1, \mathfrak{g}_1], \mathfrak{g}_2=[\mathfrak{g}_2,\mathfrak{g}_2]$ as $\mathfrak{g}_1,\mathfrak{g}_2$ are simple ideals. Thus, we have $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$.

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One possible way (and I have not been able to think of an easier way) is the following:

Note that here I will assume this to be over $\mathbb{C}$.

That semisimple Lie algebras are perfect follows as a corollary to the theorem that such a Lie algebra is a direct sum of simple ideals. It is then clear as no non-trivial quotient of such a direct sum can be abelian.

Tobias Kildetoft
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  • I do not understand “no non-trivial quotient of such a direct sum can be abelian”. I am not able to relate this with the result. –  Jan 20 '18 at 18:27
  • @cello Which part? That the claim holds, or that the claim leads to the Lie algebra being perfect? – Tobias Kildetoft Jan 20 '18 at 18:46
  • That the claim holds. –  Jan 20 '18 at 18:48
  • @cello Because any proper ideal will be a direct sum of some number of copies of those ideals (if we pick them correctly at least, to avoid issues with diagonals and so on). It might be a bit less clear than I realized while writing it, and I find your way more direct. – Tobias Kildetoft Jan 20 '18 at 18:52
  • Thanks for compliment. I would try to understand your idea spending some time. Will come back and ask you if I still have some doubt. Thank you. –  Jan 20 '18 at 19:05
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If you are allowed to use the fact that a quotient of $\mathfrak{g}$ (a finite dimension semisimple Lie algebra over a field with characteristic $0$) is semisimple, then there is an easy answer. The usual way to prove this fact is to use Cartan's criterion: see here.

If $\mathfrak{g}$ is semisimple, then its quotient $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ is also semisimple. But this is abelian, so $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}] = 0$, then $\mathfrak{g} =[\mathfrak{g},\mathfrak{g}]$.

Jianing Song
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here you can find some relevant discussion.

https://mathoverflow.net/questions/60498/lie-algebra-semisimple

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GA316
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